# Converse to the Eilenberg-Steenrod theorem?

For the purposes of this question, a homology theory is a covariant functor from the homotopy category of finite pointed CW complexes to graded abelian groups, and a collection of connecting homomorphisms, which satisfy the Eilenberg-Steenrod axioms for reduced homology, minus the dimension axiom. So this amounts to the statement that for $f:A \to X$, the connecting homomorphism and the functor grades give a long exact sequence using the cone of $f$, along with excision (where relative homology is just homology of the cone); I don't think(?) I need the suspension axiom. Dually, a cohomology theory is a contravariant functor and connecting homomorphisms satisfying the dual axioms.

The Eilenberg-Steenrod theorem, as I understand it (EDIT According to Qiaochu's second comment below, though, I understand incorrectly!), says that if two homology theories have isomorphic values on $S^0$, then this isomorphism extends uniquely to a natural isomorphism between their underlying functors, and their connecting homomorphisms agree under this isomorphism (actually this can't quite be true -- you could always replace the connecting homomorphism with its negative. I'm not sure how far from the truth this is, though.) Similarly for cohomology.

How about the converse -- given a graded abelian group $A$, does there exist a (co)homology theory which takes $S^0$ to $A$?

Brown representability isn't what I want -- that requires you to already have the functor and builds a spectrum (in fact it's almost the opposite, and it's suspicious that I seem to be replacing the role of the spectrum with a lowly graded abelian group).

It does seem to be related to the Atiyah-Hirzebruch spectral sequence. I don't imagine the spectral sequence is well-defined if I haven't actually specified a functor, but if it were, then I could attempt to piece together the functor from the output of the spectral sequence. The difficulty of doing this makes me suspect that I've got it all wrong.

My secret goal here is that I'd like a purely formal construction of ordinary homology. The most formal way I know to do this is to define it representably using an Eilenberg-Mac Lane spectrum. But it still takes some work to construct this spectrum, and I'd like to circumvent this by appealing to more general machinery.

EDIT In light of Qiaochu's answer below, there's an important lesson to be learned. The Eilenberg-Steenrod axioms including the dimension axiom are enough to calculate the ordinary (co)homology of any finite CW complex. Removing the dimension axiom gives the right notion of generalized (co)homology theory. Nevertheless, replacing the dimension axiom in the obvious way by specifying a different value for the (co)homology of $S^0$ is not enough information to determine a (co)homology functor, so the Eilenberg-Steenrod axioms + an alternate dimension axiom will not provide enough calculational rules to determine the generalized cohomology of an arbitrary CW complex.

The fact that ordinary (co)homology is determined qua (co)homology theory by its value at a point, then, is a special fact which resembles the special fact that a $K(A,n)$ is determined qua space by its homotopy groups. I wonder if there is a connection...

• I wouldn't call that the converse. Eilenberg-Steenrod answers a uniqueness question (in some sense) and you want to answer an existence question. – Qiaochu Yuan Dec 9 '14 at 6:50
• By the way, that is not what Eilenberg-Steenrod says. You need to provide a natural transformation of (co)homology theories inducing an isomorphism on $S^0$ (same as the condition in Whitehead's theorem: you need a map inducing an isomorphism, not just an isomorphism). Even when the graded abelian groups assigned to $S^0$ are isomorphic such a natural transformation need not exist, e.g. it does not exist for periodic ordinary cohomology vs. complex K-theory. So I was being imprecise: Eilenberg-Steenrod really answers a faithfulness question, not a uniqueness question. – Qiaochu Yuan Dec 9 '14 at 7:41
• Regarding your "formal construction" you might find my answer to math.stackexchange.com/questions/1055619 relevant. Maybe you are looking for something quite different. I still feel it is worth looking at the history. – Ronnie Brown Dec 9 '14 at 11:29
• @QiaochuYuan It's not quite a faithfulness question, from what you're saying -- it's that the "evaluate at $S^0$" functor reflects isomorphisms. – tcamps Dec 9 '14 at 13:30
• @tcamps: regarding the edit, the connection is that the value of a generalized cohomology theory determined by some spectrum at a point is precisely the homotopy groups of the spectrum. Like spaces, spectra are not in general determined by their homotopy groups, but they are in the Eilenberg-MacLane case, and like spaces, the difference is captured by the formalism of Postnikov towers and $k$-invariants. – Qiaochu Yuan Dec 9 '14 at 22:08

Yes. There is always the following boring choice (which I'll write as an unreduced theory): take the cohomology theory which in degree $n$ assigns

$$X \mapsto \prod_m H^{n-m}(X, A_m)$$

and similarly for homology and similarly for pairs and similarly for the reduced theory. One way to capture the difference between this boring theory and some interesting theory with the same value on a point (remember, unreduced) is the differentials in the Atiyah-Hirzebruch spectral sequence from the $E_2$ page on. For the above theory all of those differentials vanish, but in general that's not true.

My secret goal here is that I'd like a purely formal construction of ordinary homology

What counts as a purely formal construction of ordinary homology if "do the thing to the Eilenberg-MacLane spectrum" doesn't? What's another example of what you'd consider a purely formal construction?

• Certainly I'm being a bit perverse here. But constructing the Eilenberg-Mac Lane spectrum requires you to specify maps $K(A,n-1) \to \Omega K(A,n)$. The way I've seen this done is to appeal to the fact that $K(A,n)$'s are uniquely determined by their homotopy groups, and this requires some real effort to show -- it doesn't feel like a formal fact to me. It would be really nice to instead construct cohomology in some other formal way, and then show that any space with the homotopy groups of a $K(A,n)$ represents the $n$th degree of this cohomology, and appeal to Yoneda. – tcamps Dec 9 '14 at 13:37
• What I'd like to do is to treat the Eilenberg-Steenrod axioms as providing "algebraic operations and relations" under which the homology theory is "freely generated" from its value at a point, just like a free group is generated by its generators. In the $\infty$-category setting, spaces are the free $\infty$-category with $\infty$-colimits on a point. I guess I want a 1-categorical shadow of this fact, and I want the shadow of $\infty$-cocontinuous functor" to be "functor satisfying the Eilenberg-Steenrod axioms". Maybe I need to replace graded abelian groups by derived chain complexes. – tcamps Dec 9 '14 at 17:03
• @tcamps: yes, that's an option. You can replace "cocontinuous functor to spectra" with "cocontinuous functor to chain complexes" (where by chain complexes I mean the $(\infty, 1)$-category presented by chain complexes), then take homology. – Qiaochu Yuan Dec 9 '14 at 17:19
• @tcamps: you can also just write down better explicit models of the spaces in the Eilenberg-MacLane spectrum. There is an explicit model of the delooping construction $G \mapsto BG$ which, if given a topological abelian group as input, produces a topological abelian group as output. Iterating that construction starting from an abelian group $A$ produces a family of topological abelian groups $B^n A \cong K(A, n)$, and from the explicit model it shouldn't be very hard to write down explicit maps $B^{n-1} A \to \Omega B^n A$. – Qiaochu Yuan Dec 9 '14 at 17:22
• @tcamps: you may also be interested in the Dold-Thom theorem (en.wikipedia.org/wiki/Dold%E2%80%93Thom_theorem), which at least provides a definition of homology involving very little work but with which you can do very little until you prove some non-formal statements about how it behaves. – Qiaochu Yuan Dec 10 '14 at 22:24